又是一道水题= =，突然发现第一页下面好多水题，等我A了它们= =
可以发现ans=sigma(t[i]*k) k表示是第几个修，然后将m拆成m*n个点，表示第几次修，从s向n*m个点连流量为1的边，从n*m个点分别向车连权值为t[i]*k的边，然后再从车向T连边就行了（和noi的某道题一样还有GDKOI2014T2一样）
CODE：
#include<cstdio>#include<iostream>#include<cstring>#include<algorithm>#include<queue>#define maxn 650#define maxm 180000#define inf 0x7fffffffusing namespace std;
struct edges{
int to,cap,dist,next;
}edge[maxm];int s,t,next[maxn],l;
int addedge(int from,int to,int cap,int dist){
l++;
edge[l*2]=(edges){to,cap,dist,next[from]};
edge[l*2+1]=(edges){from,0,-dist,next[to]};
next[from]=l*2;next[to]=l*2+1;
return 0;
}bool b[maxn];
int dist[maxn],way[maxn];
queue<int> q;
bool spfa(){
for (int i=1;i<=t;i++) dist[i]=inf;
memset(b,0,sizeof(b));
dist[s]=0;
q.push(s);
while (!q.empty()){
int u=q.front();q.pop();
b[u]=0;
for (int i=next[u];i;i=edge[i].next)
if (edge[i].cap&&dist[u]+edge[i].dist<dist[edge[i].to]) {
dist[edge[i].to]=dist[u]+edge[i].dist;
way[edge[i].to]=i;
if (!b[edge[i].to]){
b[edge[i].to]=1;q.push(edge[i].to);
}
}
}
if (dist[t]==inf) return 0;
return 1;
}int mcmf(){
int cost=0;
while (spfa()){
cost+=dist[t];
int x=t;
while (x!=s){
edge[way[x]].cap-=1;
edge[way[x]^1].cap+=1;
x=edge[way[x]^1].to;
}
}
return cost;
}int n,m;
int main(){
scanf("%d%d",&m,&n);
s=n*m+n+1;t=n*m+n+2;
for (int i=1;i<=n;i++)
for (int j=1;j<=m;j++){
int x;
scanf("%d",&x);
for (int k=1;k<=n;k++) addedge((k-1)*m+j,n*m+i,1,x*k);
}
for (int i=1;i<=n*m;i++) addedge(s,i,1,0);
for (int i=1;i<=n;i++) addedge(n*m+i,t,1,0);
printf("%.2lf\n",mcmf()*1.0/n);
return 0;
} |